Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 8x}{x + 8} = \dfrac{-3x - 24}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 8x}{x + 8} (x + 8) = \dfrac{-3x - 24}{x + 8} (x + 8)$ $ x^2 + 8x = -3x - 24$ Subtract $-3x - 24$ from both sides: $ x^2 + 8x - (-3x - 24) = -3x - 24 - (-3x - 24)$ $ x^2 + 8x + 3x + 24 = 0$ $ x^2 + 11x + 24 = 0$ Factor the expression: $ (x + 3)(x + 8) = 0$ Therefore $x = -3$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.